|
%I #50 Mar 21 2024 08:35:09
%S 1,5,5,25,125,3125,390625,1220703125,476837158203125,
%T 582076609134674072265625,277555756156289135105907917022705078125,
%U 161558713389263217748322010169914619837072677910327911376953125
%N a(n) = a(n-1)*a(n-2) with a(0)=1, a(1)=5.
%C a(17) has 1117 digits.
%C From _Peter Bala_, Nov 01 2013: (Start)
%C Let phi = 1/2*(1 + sqrt(5)) denote the golden ratio A001622. This sequence is the simple continued fraction expansion of the constant c := 4*Sum_{n = 1..oo} 1/5^floor(n*phi) (= 16*Sum_{n = 1..oo} floor(n/phi)/5^n) = 0.83866 83869 91037 14262 ... = 1/(1 + 1/(5 + 1/(5 + 1/(25 + 1/(125 + 1/(3125 + 1/(390625 + ...))))))). The constant c is known to be transcendental (see Adams and Davison 1977). Cf. A014565.
%C Furthermore, for k = 0,1,2,... if we define the real number X(k) = sum {n >= 1} 1/5^(n*Fibonacci(k) + Fibonacci(k+1)*floor(n*phi)) then the real number X(k+1)/X(k) has the simple continued fraction expansion [0; a(k+1), a(k+2), a(k+3), ...] (apply Bowman 1988, Corollary 1). (End)
%H Vincenzo Librandi, <a href="/A214706/b214706.txt">Table of n, a(n) for n = 0..16</a>
%H W. W. Adams and J. L. Davison, <a href="http://www.jstor.org/stable/2041889">A remarkable class of continued fractions</a>, Proc. Amer. Math. Soc. 65 (1977), 194-198.
%H P. G. Anderson, T. C. Brown, and P. J.-S. Shiue, <a href="http://people.math.sfu.ca/~vjungic/tbrown/tom-28.pdf">A simple proof of a remarkable continued fraction identity</a>, Proc. Amer. Math. Soc. 123 (1995), 2005-2009.
%H D. Bowman, <a href="https://www.fq.math.ca/Scanned/26-1/bowman.pdf">A new generalization of Davison's theorem</a>, Fib. Quart. Volume 26 (1988), 40-45.
%F a(n) = 5^Fibonacci(n).
%p a:= n-> 5^(<<1|1>, <1|0>>^n)[1, 2]:
%p seq(a(n), n=0..12); # _Alois P. Heinz_, Jun 17 2014
%t 5^Fibonacci[Range[0,11]]
%t nxt[{a_,b_}]:={b,a*b}; NestList[nxt,{1,5},12][[All,1]] (* _Harvey P. Dale_, Oct 14 2018 *)
%o (Magma) [5^Fibonacci(n): n in [0..13]];
%o (SageMath) [5^fibonacci(n) for n in range(15)] # _G. C. Greubel_, Jan 07 2024
%Y Cf. A000045, A000301, A001622, A010098, A010099, A010100, A014565, A214706, A214887, A215270, A215271, A215272.
%Y Column k=5 of A244003.
%K nonn,easy
%O 0,2
%A _Vincenzo Librandi_, Aug 01 2012
|
